1. \( P(X)=\frac{1}{5} \), for \( X=0,1,2,3,4,5 \)

It seems like you are providing a probability distribution function \( P(X) \) for a discrete random variable \( X \) that takes on values from 0 to 5. The function is defined as \( P(X) = \frac{1}{5} \) for each of these values. Let's analyze this step by step: 1. **Identify the values of \( X \)**: The random variable \( X \) can take the values \( 0, 1, 2, 3, 4, 5 \). 2. **Probability Distribution**: The probability for each value of \( X \) is given as \( P(X) = \frac{1}{5} \). This means that each outcome has an equal probability of occurring. 3. **Check the total probability**: The sum of the probabilities for all possible values of \( X \) should equal 1. We can calculate this as follows: \[ P(0) + P(1) + P(2) + P(3) + P(4) + P(5) = \frac{1}{5} + \frac{1}{5} + \frac{1}{5} + \frac{1}{5} + \frac{1}{5} + \frac{1}{5} \] Calculating this gives: \[ 6 \times \frac{1}{5} = \frac{6}{5} \] Since \( \frac{6}{5} \) is greater than 1, this indicates that the probabilities do not sum to 1, which violates the fundamental property of probability distributions. ### Conclusion The provided probability distribution \( P(X) = \frac{1}{5} \) for \( X = 0, 1, 2, 3, 4, 5 \) is incorrect because the total probability exceeds 1. To correct this, the probabilities should be adjusted so that they sum to 1. Since there are 6 outcomes, each probability should be: \[ P(X) = \frac{1}{6} \] for \( X = 0, 1, 2, 3, 4, 5 \). If you have any further questions or need additional calculations, feel free to ask!